Organic Chemistry
Organic Chemistry
3rd Edition
ISBN: 9781119316152
Author: Klein, David R.
Publisher: WILEY
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Chapter 1, Problem 50PP

(a)

Interpretation Introduction

Interpretation:

The geometry of the central atoms in NH3 has to be predicted.

Concept Introduction:

According to VSEPR (Valence Shell Electron Pair Repulsion) theory, each molecule gets a unique structure. That structure is explained by considering steric number of that molecule.

The steric number is the combination of both number of σ -bonds and number of lone pairs involved in a particular molecule.

The geometry of the central atom will be determined by counting the steric number followed by the hybridization state of that central atom and finally electronic arrangement of atoms in space.

If the steric number is 4, the central atom has sp3 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be tetrahedral.

If the steric number is 3, the central atom has sp2 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be trigonal planar.

If the steric number is 2, the central atom has sp hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be linear.

sp3 hybridized =    tetrahedral

    trigonal pyramidal(if one lone pair of electron is present)

    bent geometry (if two lone pair of electrons are present)

sp2 hybridized =    trigonal planar

sp hybridized =    linear

(b)

Interpretation Introduction

Interpretation:

The geometry of the central atoms in BH3 has to be predicted.

Concept Introduction:

According to VSEPR (Valence Shell Electron Pair Repulsion) theory, each molecule gets a unique structure. That structure is explained by considering steric number of that molecule.

The steric number is the combination of both number of σ -bonds and number of lone pairs involved in a particular molecule.

The geometry of the central atom will be determined by counting the steric number followed by the hybridization state of that central atom and finally electronic arrangement of atoms in space.

If the steric number is 4, the central atom has sp3 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be tetrahedral.

If the steric number is 3, the central atom has sp2 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be trigonal planar.

If the steric number is 2, the central atom has sp hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be linear.

sp3 hybridized =    tetrahedral

    trigonal pyramidal(if one lone pair of electron is present)

    bent geometry (if two lone pair of electrons are present)

sp2 hybridized =    trigonal planar

sp hybridized =    linear

(c)

Interpretation Introduction

Interpretation:

The geometry of the central atoms in CH3+ has to be predicted.

Concept Introduction:

According to VSEPR (Valence Shell Electron Pair Repulsion) theory, each molecule gets a unique structure. That structure is explained by considering steric number of that molecule.

The steric number is the combination of both number of σ -bonds and number of lone pairs involved in a particular molecule.

The geometry of the central atom will be determined by counting the steric number followed by the hybridization state of that central atom and finally electronic arrangement of atoms in space.

If the steric number is 4, the central atom has sp3 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be tetrahedral.

If the steric number is 3, the central atom has sp2 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be trigonal planar.

If the steric number is 2, the central atom has sp hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be linear.

sp3 hybridized =    tetrahedral

    trigonal pyramidal(if one lone pair of electron is present)

    bent geometry (if two lone pair of electrons are present)

sp2 hybridized =    trigonal planar

sp hybridized =    linear

(d)

Interpretation Introduction

Interpretation:

The geometry of the central atoms in CH3 has to be predicted.

Concept Introduction:

According to VSEPR (Valence Shell Electron Pair Repulsion) theory, each molecule gets a unique structure. That structure is explained by considering steric number of that molecule.

The steric number is the combination of both number of σ -bonds and number of lone pairs involved in a particular molecule.

The geometry of the central atom will be determined by counting the steric number followed by the hybridization state of that central atom and finally electronic arrangement of atoms in space.

If the steric number is 4, the central atom has sp3 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be tetrahedral.

If the steric number is 3, the central atom has sp2 hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be trigonal planar.

If the steric number is 2, the central atom has sp hybridized and the electronic arrangement of atoms in space (i.e. geometry) will be linear.

sp3 hybridized =    tetrahedral

    trigonal pyramidal(if one lone pair of electron is present)

    bent geometry (if two lone pair of electrons are present)

sp2 hybridized =    trigonal planar

sp hybridized =    linear

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Chapter 1 Solutions

Organic Chemistry

Ch. 1.3 - Prob. 8PTSCh. 1.3 - Prob. 9PTSCh. 1.3 - Prob. 10PTSCh. 1.3 - Prob. 11ATSCh. 1.4 - Prob. 4LTSCh. 1.4 - Prob. 12PTSCh. 1.4 - Prob. 13PTSCh. 1.4 - Prob. 14ATSCh. 1.5 - Prob. 5LTSCh. 1.5 - Prob. 15PTSCh. 1.5 - Prob. 16ATSCh. 1.5 - Prob. 17ATSCh. 1.6 - Prob. 6LTSCh. 1.6 - Prob. 18PTSCh. 1.6 - Prob. 19PTSCh. 1.6 - Prob. 20ATSCh. 1.9 - Prob. 21CCCh. 1.9 - Prob. 22CCCh. 1.9 - Prob. 23CCCh. 1.9 - Prob. 7LTSCh. 1.9 - PRACTICE the skill Determine the hybridization...Ch. 1.9 - APPLY the skill Nemotin is a compound that was...Ch. 1.9 - Prob. 26CCCh. 1.10 - 1.8 PREDICTING GEOMETRY LEARN the skill Using...Ch. 1.10 - PRACTICE the skill Use VSEPR theory to Predict the...Ch. 1.10 - Prob. 28PTSCh. 1.10 - Ammonia (NH3) will react with a strong acid, such...Ch. 1.10 - Volatile organic compounds (VOCs) contribute to...Ch. 1.11 - Prob. 9LTSCh. 1.11 - Prob. 31PTSCh. 1.11 - Volatile organic compounds (VOCs) contribute to...Ch. 1.12 - Prob. 10LTSCh. 1.12 - Prob. 33PTSCh. 1.12 - Epichlorohydrin (1) is an epoxide used in the...Ch. 1 - Prob. 35PPCh. 1 - Prob. 36PPCh. 1 - Prob. 37PPCh. 1 - Prob. 38PPCh. 1 - Prob. 39PPCh. 1 - Prob. 40PPCh. 1 - Prob. 41PPCh. 1 - Prob. 42PPCh. 1 - Prob. 43PPCh. 1 - Prob. 44PPCh. 1 - Prob. 45PPCh. 1 - Prob. 46PPCh. 1 - Prob. 47PPCh. 1 - Prob. 48PPCh. 1 - Prob. 49PPCh. 1 - Prob. 50PPCh. 1 - Prob. 51PPCh. 1 - Prob. 52PPCh. 1 - Prob. 53PPCh. 1 - Prob. 54PPCh. 1 - Prob. 55PPCh. 1 - Prob. 56PPCh. 1 - Prob. 57PPCh. 1 - Prob. 58PPCh. 1 - Prob. 59PPCh. 1 - Prob. 60PPCh. 1 - Prob. 61PPCh. 1 - Prob. 62PPCh. 1 - Prob. 63PPCh. 1 - Prob. 64PPCh. 1 - Prob. 65PPCh. 1 - Prob. 66IPCh. 1 - Propose at least two different structures for a...Ch. 1 - Prob. 68IPCh. 1 - Prob. 69IPCh. 1 - Prob. 70IPCh. 1 - Prob. 71IPCh. 1 - Prob. 72IPCh. 1 - Prob. 73IPCh. 1 - Prob. 74IPCh. 1 - Prob. 75IPCh. 1 - Prob. 76IPCh. 1 - Prob. 77IPCh. 1 - Prob. 78CPCh. 1 - Prob. 79CPCh. 1 - Prob. 80CPCh. 1 - Prob. 81CP
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